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Chapter 9: Problem 1321

The relation between two specific heats of a gas is (A) $\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=(\mathrm{R} / \mathrm{J})$ (B) \(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{J}\) (C) $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=(\mathrm{R} / \mathrm{J})$ (D) \(\mathrm{C}_{\mathrm{V}}-\mathrm{C}_{\mathrm{P}}=\mathrm{J}\)

Short Answer

Expert verified
The correct relation between the specific heats of a gas, Cp and Cv, is given by Mayer's formula: \(C_p - C_v = R\). However, none of the provided options represent the correct Mayer's formula.

Step by step solution

01

Option A: Cv - Cp = (R / J)

Cv and Cp are specific heats at constant volume and constant pressure, respectively. Subtracting Cv from Cp in this manner (Cv - Cp) does not give the correct relationship as per Mayer's formula. Hence, option A is incorrect.
02

Option B: Cp - Cv = J

Subtracting specific heats at constant volume from specific heats at constant pressure (Cp - Cv) is closer to the correct relation. However, the right side should have "R" instead of "J." Therefore, option B is also incorrect.
03

Option C: Cp - Cv = (R / J)

According to Mayer's formula, the difference between specific heats at constant pressure and constant volume is equal to the gas constant R. Here, the right side is given by (R / J), which is not the correct relation. Hence, option C is incorrect.
04

Option D: Cv - Cp = J

As discussed in option A, subtracting Cv from Cp in this manner does not give the correct relationship. The right side should have "R," not "J". Thus, option D is also incorrect. However, after analyzing all of the given options and none of them seem to represent the correct Mayer's formula. The correct Mayer's formula should be: Cp - Cv = R Unfortunately, none of the options given are correct.

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Most popular questions from this chapter

When the pressure on \(1200 \mathrm{ml}\) of a gas is increased from $70 \mathrm{~cm}\( to \)120 \mathrm{~cm}$ of mercury at constant temperature, the new volume of the gas will be (A) \(400 \mathrm{ml}\) (B) \(600 \mathrm{ml}\) (C) \(700 \mathrm{ml}\) (D) \(500 \mathrm{ml}\)

A gas at \(27^{\circ} \mathrm{C}\) has a volume \(\mathrm{V}\) and pressure \(\mathrm{P}\). On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be (A) \(1527^{\circ} \mathrm{C}\) (B) \(600^{\circ} \mathrm{C}\) (C) \(162^{\circ} \mathrm{C}\) (D) \(1800^{\circ} \mathrm{C}\)

The molecules of a given mass of a gas have a rms velocity of $200 \mathrm{~m} / \mathrm{s}\( at \)27^{\circ} \mathrm{C}\( and \)1.0 \times 10^{5} \mathrm{Nm}^{-2}\( pressure when the temperature is \)127^{\circ} \mathrm{C}$ and pressure is \(0.5 \times 10^{5} \mathrm{Nm}^{-2}\), the rms velocity in \(\mathrm{m} / \mathrm{s}\) will be (A) \(100 \sqrt{2}\) (B) \([(100 \sqrt{2}) / 3]\) (C) \([(400) / \sqrt{3}]\) (D) None of these

When the temperature of a gas is raised from \(27^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\), the percentage increase in the rms velocity of the molecules will be (A) \(15 \%\) (B) \(17.5 \%\) (C) \(10 \%\) (D) \(20 \%\)

\(\mathrm{O}_{2}\) gas is filled in a vessel. If pressure is double, temperature becomes four times, how many times its density will become. (A) 4 (B) \((1 / 4)\) (C) 2 (D) \((1 / 2)\)
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